172. Factorial Trailing Zeroes*

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本文介绍了一种高效算法，用于计算给定整数n的阶乘末尾零的数量。该算法采用递归方式，实现了对数时间复杂度，特别适用于大数据计算场景。

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<span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; background-color: rgb(255, 255, 255);">Given an integer </span><span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; background-color: rgb(255, 255, 255);">n</span><span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; background-color: rgb(255, 255, 255);">, return the number of trailing zeroes in </span><span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; background-color: rgb(255, 255, 255);">n</span><span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; background-color: rgb(255, 255, 255);">!.</span>

Note: Your solution should be in logarithmic time complexity.

Reference

<pre name="code" class="python">class Solution(object):
    def trailingZeroes(self, n):
        """
        :type n: int
        :rtype: int
        """
        return 0 if n == 0 else n / 5 + self.trailingZeroes(n / 5)

</pre><pre name="code" class="python">class Solution(object):
    def trailingZeroes(self, n):
        """
        :type n: int
        :rtype: int
        """
        result = 0
        while n>=5:
            n/=5
            result +=n
        return result